Tuesday
July 22, 2014

Homework Help: Geometry

Posted by gr8 on Monday, January 15, 2007 at 5:03pm.

An isosceles right triangle has an area of 98cm squared. Find the length of each leg.

The formula of finding the area of a triangle with one right angle is:

(b x h)/2

You know that an isoceles triangles has two equal sides. Those equal sides are the base and height. This means that both the base height with have the same value.

After you get the answer of the two isoceles legs, you need to figure out the other leg. To solve this by using the formula:

a^2+b^2=c^2

a would be on leg length
b would be the other leg length

You just fill in the values and you get your answer. =)


this did not help

If the area is 98 then the sides would be equal to b x h / 2. So they would be

98=b x h /2
98 x 2= b x h
196= b x h (we know that the base and the height are the same so you can just find the square root of 196)

square root of 196= 14x14

So the two sides are 14 and 14

Now we need to find the third side.

We will use the formula:

a^2+b^2=c^2

14^2+14^2=c^2
196+196=c^2
329=c^2

So the side lengths of the triangle are

(14,14, square root of 329)



f

Answer this Question

First Name:
School Subject:
Answer:

Related Questions

Math (Geometry) - The perimeter of right triangle ABC is equal to the perimeter ...
Math - 1) The hypotenuse of an isosceles right triangle is 8 cm longer than ...
math - i have trouble finding the answer to this and how to srt. there is a ...
MATH MS.SUE HELP PLEASE! - Could somebody help me I don't understand any of this...
Math - "One leg of a right triangle is 50.A line parallel to the other leg and ...
Geometry - Right triangle ABC has one leg of length 12 in and a hypotenuse of ...
geometry - The perimeter of right triangle RST is equal to the perimeter of ...
geometry - Each base of an isosceles triangle measures 42 degrees, 30'. The base...
geometry - leg xy of the right triangle is twice as long as leg yz. if the area ...
Math...Geometry - Hello, Can someone please tell me if I worked this out correct...

Search
Members